Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200006;
int sy, L, R, n, q, sum[maxn];
struct data {
  int l, r;
  bool operator<(const data& b) const { return l < b.l; }
} a[maxn];
double cal(int x, int y, int Mx) {
  return (double)Mx * (1 - (double)sy / y) + (double)sy / y * x;
}
int findl(int x, int y) {
  int l = 1, r = n;
  while (l <= r) {
    int mid = l + r >> 1;
    if (cal(x, y, a[mid].r) > L)
      r = mid - 1;
    else
      l = mid + 1;
  }
  return l;
}
int findr(int x, int y) {
  int l = 1, r = n;
  while (l <= r) {
    int mid = l + r >> 1;
    if (cal(x, y, a[mid].l) < R)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
int main() {
  scanf("%d%d%d%d", &sy, &L, &R, &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &a[i].l, &a[i].r);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i].r - a[i].l;
  scanf("%d", &q);
  while (q--) {
    int x, y, t1, t2;
    double ans;
    scanf("%d%d", &x, &y);
    t1 = findl(x, y);
    t2 = findr(x, y);
    if (t1 == t2) {
      ans = min((double)R, cal(x, y, a[t1].r)) -
            max((double)L, cal(x, y, a[t1].l));
      printf("%.10lf\n", ans);
      continue;
    }
    if (t1 > t2) {
      ans = 0;
      printf("%.10lf\n", ans);
      continue;
    }
    ans = (double)(sum[t2 - 1] - sum[t1]) * (y - sy) / y;
    if (cal(x, y, a[t1].r) > L)
      ans += cal(x, y, a[t1].r) - max((double)L, cal(x, y, a[t1].l));
    if (cal(x, y, a[t2].l) < R)
      ans += min((double)R, cal(x, y, a[t2].r)) - cal(x, y, a[t2].l);
    printf("%.10lf\n", ans);
  }
  return 0;
}